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2 years ago

Tina says that 3.4 x 105 is greater than 8.25 x 104. Is she correct? Justify your answer.

Mathematics

2 answers:

Anna71 [15]2 years ago

6 0

Answer:

No, Tina is not correct since 3.4 × 105 = 357 and 8.25 × 104 = 858

Umnica [9.8K]2 years ago

4 0

No that is not correct 3.4x105=425 and 8.25x104=858

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13 kids sandwiches X 4 inches = 52 inches
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17 medium sandwiches X 8 inches = 136 inches
17 large sandwiches X 12 inches = 204 inches

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3 years ago

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iogann1982 [59]

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3 years ago

Read 2 more answers

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = the number that are not correctly calibrated.

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = 0.35

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = 0.54

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = 0.11

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago

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Step2247 [10]

Answer:

Step-by-step explanation:

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2+3

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3 years ago

Answer all questions shown

FromTheMoon [43]

SOH CAH TOA

—>

Sin D = 13 sqrt3/26 = 0.76
Sin E = 13/26 = 0.47
Cos D = 13/26 = 0.87
Cos E = 13 sqrt3/26 = 0.64

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2 years ago